奥本海姆信号与系统上册2版课后答案

1、易考网 www.ekaokao.com 考研真题 |课后答案 全部免费 第一章 1.3 解： (a). 2401lim ( ) , 04TtTTExtdtedP=(b) dttxTPTTT=2)(21lim 121lim =dtTTTT=dttxdttxETTT22)()(lim (c). 222lim ( ) cos ( ) ,111cos(2)1lim ( ) lim22 2TTTTTTTExtdttdtPxtd d =+= =(d) 034121lim)21(121lim121lim022=+=+=+= NNnxNPNNnnNNNnN34)21()(lim202=nNNnNnxE (e). 2() 1,xn E= 211lim lim 1 121 21NNNNnN nNPxn = =+(f) =+=NNnNnxNP21)(121lim2=NNnNn。

2、 Signals & Systems (Second Edition) Learning Instructions (Exercises Answers) Department of Computer Engineering 2005.12 1 Contents Chapter 1 2 Chapter 2
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3、158Chapter 9 Answers9.1 （a）The given integral may be written as (5)0tjedIf -5,then the integral does converge (b) The given integral may be written ast0(5)tjeIf -5 ,then the function grows towards as t decreases towards - and the given integral does not (5)te converge .but if -5 ,then the function grows towards as t decreases towards - and the given integral (5)tedoes not converge If -5,therefore, the given internal converges when 5 ,then the function grows towards as t de。

4、Signals & Systems(Second Edition)Learning Instructions(Exercises Answers)Department of Computer Engineering 2005.121ContentsChapter 1 2Chapter 2 &。

5、80Chapter 4 Answers4.1 (a)Let then the Fourier transform is :.12tutxt tofxj jdtdttjXjjt jt 2/11122is as shown in figure s4.1.j(b) Let then the Fourier transform is :.12ttx tofxjX21124/ jjj tttttt dddXis as shown in figure s4.1j4.2 (a) Let then the Fourier transform is :.11ttxtofxjX1cs2jjtjdjXis as sketched in figure s4.2.j1(b)the signal is as shown in the figure below .Clearly,22tuttudTherefore )2sin(。

6、17Chapter 2 answers 2.1 (a) We have know that 1*kyxnhxnThe signals xn and hn are as how in Figure S2.1From this figure, we can easily see that the above convolution sum reduces to11ynhxhxn2This gives14224n(b)We know that2*kynxhnxnkComparing with eq.(S2.1-1),we see that21y(c) We may rewrite eq.(S2.1-1) as1*kynxhxhnkSimilarly, we may write322kComparing this with eq.(S2.1),we see that31yn2.2 Using given definition for the signal hn, w。